What is the orbit of 1 in the permutation group G?

[mathplease]

If G is a permutation group acting on a finite set A and x \inA, then

|G| = |\Delta(x)| |G_x|

where \Delta(x) denotes the orbit containing x.

I'm having some difficulty understanding this theorem. For example,

Consider the following set G of permutations of the set M = {1,2,3,4}:

* e = (1)(2)(3)(4)

* a = (1 2)(3)(4) = (1 2)

* b = (1)(2)(3 4) = (3 4)

* ab = (1 2)(3 4)

G forms a group, so (G,M) forms a permutation group.

what is \Delta(1) in this example?

 

[Fredrik]

I'm having some difficulty understanding your notation. Can you explain it?

 

[mathplease]

I'm having some difficulty understanding your notation. Can you explain it?

|A| means order of A

G is the the permutation group

\Delta(x) is the http://en.wikipedia.org/wiki/Orbit_(group_theory)#Orbits_and_stabilizers" containing x (having difficulty understanding/visualising this)

G_x is all the permutations in G that fix x (map x to itself) also known as the stabiliser of x in G

also I'm using http://en.wikipedia.org/wiki/Cycle_notation#Definition" for the permutations:

(1 2)(3 4) means 1 -> 2, 2 -> 1, 3 -> 4, 4 -> 3

 

[Landau]

The orbit of an element is just the set of images under the action.

e1=1
a1=2
b1=1
(ab)1=2

So the orbit of 1 is {1,2}, which is also the orbit of 2. Similarly, the orbit of 3 is the set {3,4}, which is also the orbit of 4. This illustrates the fact that the orbits form a partition of the G-set.

 

[mathplease]

The orbit of an element is just the set of images under the action.

e1=1
a1=2
b1=1
(ab)1=2

So the orbit of 1 is {1,2}.

Thankyou very much!